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5-47E

Expert-verifiedFound in: Page 321

Book edition
13th

Author(s)
James T. McClave, P. George Benson, Terry Sincich

Pages
888 pages

ISBN
9780134506593

**Question: Hotel guest satisfaction. Refer to the results of the 2015 North American Hotel Guest Satisfaction Index Study, Exercise 4.49 (p. 239). Recall that 15% of hotel guests were “delighted” with their experience (giving a rating of 10 out of 10); of these guests, 80% stated they would “definitely” recommend the hotel. In a random sample of 100 hotel guests, find the probability that fewer than 10 were delighted with their stay and would recommend the hotel.**

The probability that fewer than 10 () delighted with their stay and would recommend the hotel is 0.2676.

Let *p* denotes the proportion of guests who were delighted with their stay and would recommend the hotel.

15% of guests were delighted with their experience, and of these, 80% would recommend the hotel

Therefore,

.

A random sample of size 100 is selected.

The probability that fewer than 10 () delighted with their stay and would recommend the hotel obtain as follows

$P\left(\hat{p}<.10\right)=P\left(\frac{\hat{p}-p}{{\sigma}_{\hat{p}}}<\frac{0.10-p}{{\sigma}_{\hat{p}}}\right)\phantom{\rule{0ex}{0ex}}=P\left(Z<\frac{0.10-0.12}{\sqrt{\frac{0.12\times 0.88}{100}}}\right)\phantom{\rule{0ex}{0ex}}=P\left(Z<\frac{-0.02}{\sqrt{0.001056}}\right)\phantom{\rule{0ex}{0ex}}=P\left(Z<\frac{-0.02}{0.0324}\right)\phantom{\rule{0ex}{0ex}}=P\left(Z<-0.62\right)\phantom{\rule{0ex}{0ex}}=0.2676\phantom{\rule{0ex}{0ex}}$In the z-table, the value at the intersection of -0.60 and 0.02 is the required probability.

Therefore, the required probability is 0.2676.

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